Question

An ideal gas is expanded from $p_1,V_1,T_1$ to $p_2,V_2,T_2$ under different conditions. The incorrect statement among the following is:

• A. The work done by the gas is less when it is expanded reversibly from $V_1$ to $V_2$ under adiabatic conditions as compared to that when expanded reversibly from $V_1$ to $V_2$ under isothermal conditions
• B. The change in internal energy of the gas is (i) zero, if it is expanded reversibly with $T_1=T_2$ and (ii) positive, if it is expanded reversibly under adiabatic conditions with $T_1\ne T_2$
• C. Temperature of any system in thermal equilibrium is an example of intensive property
• D. For an isothermal free expansion of an ideal gas into vacuum, $\Delta U=0,q=0,w=0$

Answer 1

The correct option is B The change in internal energy of the gas is (i) zero, if it is expanded reversibly with $T_1=T_2$ and (ii) positive, if it is expanded reversibly under adiabatic conditions with $T_1\ne T_2$

(a)
Maximum work is done on the system when compression occur irreversibly and minimum work is done in reversible compression.



$AB$ is isothermal and $AC$ is adiabatic path. Work done is area under the curve. Hence, less work is obtained in adiabatic process than in isothermal.


(b) The change in internal energy of the gas is (i) zero, if it is expanded reversibly with $T_1=T_2$ and (ii) negative, if it is expanded reversibly under adiabatic conditions with $T_1\ne T_2$. Hence, (b) is incorrect.
(c) An intensive property is a physical quantity whose value does not depend on the amount of the substance for which it is measured. For example, the temperature of a system in thermal equilibrium is the same as the temperature of any part of it.
Hence, (c) is correct.
(d)
For isothermal process, $\Delta T=0$
For an ideal gas, the internal energy $U$ depends only on temperature, not on pressure or volume.
So, $\Delta U=0$
On the basis of mathematical form of first law
$\Delta U=q+w\Rightarrow q=-w=-p\Delta V$
For an isothermal free expansion of an ideal gas into vacuum.
$p=0\therefore q=-w=0$
So, the correct value is
$\Delta U=0,q=0,w=0$